Question

determine: a) the capacitance? b) the maximum voltage that can be applied to a parallel plate capacitor filled with teflon, which has a plate area of ​​1.75 cm ^ 2 and plate separation of 0.040 mm?

Answer 1

a). Capacitance `C` of a parallel plate capacitor is given by the expression

`C = epsilon A / d` .........(1)
where `ε` is the permittivity of the dielectric material, `A` is the plate area and `d` is the plate separation.

Relative permittivity for teflon `ε_r = 2.1=>epsilon=epsilon_repsilon_0=2.1xx (8.854xx 10^-12)\ F m^-1`

Inserting various values in (1) in SI units we get

`C = 2.1 xx (8.854 xx 10^-12) xx (1.75 xx 10^-4) / (0.040 xx 10^-3)`
`=>C= 8.13 xx 10^-11\ F`

b). Dielectric strength of Teflon is `60\ MVm^-1`. As the parallel plates are `0.040\ mm` apart, the voltage difference `V` between the plates for this voltage is

`V = 60 xx 10^6 xx 0.040xx10^-3 = 2.4\ kV`